Download App

Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals4 Marks
Question
$\int\frac{3\text{x}+5}{\sqrt{7\text{x}+9}}\text{dx}$
✓

Answer

$\text{Let I}=\int\Big(\frac{3\text{x}+5}{\sqrt{7\text{x}+9}}\Big)\text{dx}$
$\text{Putting}\ 7\text{x}+9=\text{t}$
$\Rightarrow\text{x}=\frac{\text{t}-9}{7}\ \&\ 7\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{7}$
$\therefore\text{I}=\int\Bigg(\frac{3\big(\frac{\text{t}-9}{7}\big)+5}{\sqrt{\text{t}}}\Bigg)\text{dt}$
$=\int\Big(\frac{3}{7}\frac{\text{t}}{\sqrt{\text{t}}}-\frac{27}{7\sqrt{\text{t}}}+\frac{5}{\sqrt{\text{t}}}\Big)\frac{\text{dt}}{7}$
$=\frac{3}{7\times7}\int\text{t}^\frac{1}{2}\text{dt}-\frac{27}{7\times7}\int\text{t}^{-\frac{1}{2}}\text{dt}+\frac{5}{7}\int^{-\frac{1}{2}}\text{dt}$
$=\frac{3}{7\times7}\bigg[\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\bigg]-\frac{27}{7\times7}\bigg[\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\bigg]+\frac{5}{7}\bigg[\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\bigg]+\text{C}$
$=\frac{2}{7\times7}\text{t}^\frac{3}{2}-\frac{27}{7\times7}2\text{t}^\frac{1}{2}+\frac{10\sqrt{t}}{7}+\text{C}$
$=\frac{2}{7\times7}(7\text{x}+9)^\frac{3}{2}-\frac{54}{7\times7}(7\text{x}+9)^\frac{1}{2}+\frac{10}{7}\sqrt{7\text{x}+9}+\text{C}$ $[\because\text{t}=7\text{x}+9]$
$=\frac{2}{7\times7}(7\text{x}+9)^\frac{3}{2}+\Big(10-\frac{54}{7}\Big)\frac{\sqrt{7\text{x}+9}}{7}+\text{C}$
$=\frac{2}{7\times7}(7\text{x}+9)^\frac{3}{2}+\Big(\frac{70-54}{7}\Big)\frac{\sqrt{7\text{x}+9}}{7}+\text{C}$
$=\frac{2}{7\times7}(7\text{x}+9)^\frac{3}{2}+\frac{16}{7\times7}\frac{\sqrt{7\text{x}+9}}{7}+\text{C}$
$=\frac{2}{7\times7}\Big[(7\text{x}+9)^\frac{1}{2}[7\text{x}+9+8]\Big]+\text{C}$
$=\frac{2}{49}\Big[(7\text{x}+9)^\frac{1}{2}[7\text{x}+17]\Big]+\text{C}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "4 Marks" from Indefinite Integrals→Indefinite Integrals — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Differentiate the following functions with respect to x:
$\text{x}^{\frac{1}{\text{x}}}$
→
$\int\frac{2\text{x}+1}{\sqrt{3\text{x}+2}}\text{dx}$
→
In the following, determine the values of constants involved in the definition so that the given function is continuous:
$\text{f(x)}=\begin{cases}\frac{\text{k}\cos\text{x}}{\pi-2\text{x}},&\text{x}<\frac{{\pi}}{2}\\3,&\text{x}=\frac{\pi}{2}\\\frac{3\tan\text{x}}{2\text{x}-\pi},&\text{x}>\frac{\pi}{2}\end{cases}$
→
Find the area of the region included between the parabola y2 = x and the line x + y = 2.
→
Solve the following differential equation:
$\text{xy}\log\Big(\frac{\text{y}}{\text{x}}\Big)\text{dx}+\Big\{\text{y}^2-\text{x}^2\log\Big(\frac{\text{y}}{\text{x}}\Big)\Big\}\text{dy}=0$
→
Solve $\frac{\text{dy}}{\text{dx}}=\cos(\text{x}+\text{y})+\sin(\text{x}+\text{y}).$
[Hint: Substitute x + y = z]
→
If $\text{A}=\begin{bmatrix}0&\text{c}&-\text{b}\\-\text{c}&0&\text{a}\\\text{b}&-\text{a}&0\end{bmatrix}$ and $\text{B}=\begin{bmatrix}\text{a}^2&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2&\text{bc}\\\text{ac}&\text{bc}&\text{c}^2\end{bmatrix},$ show that AB = BA = O3 × 3
→
Differentiate the following functions with respect to x:
$\frac{\text{x}^2(1-\text{x}^2)}{\cos2\text{x}}$
→
Show that $\text{A}=\begin{bmatrix} 6 & 5 \\ 7 & 6 \end{bmatrix}$ satisfies the equation x2 - 12x + 1 = 0. Thus, find A-1.
→
Given the probability that A can solve a problem is $\frac{2}{3}$ and the probability that B can solve the same problem is $\frac{3}{5}$. Find the probability that none of the two will be able to solve the problem.
→